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Question Number 128707 by mnjuly1970 last updated on 09/Jan/21

...nice calculus... prove that:: ∫_0 ^( ∞) ((ln(1+ϕ^2 x^2 ))/(1+π^2 x^2 )) dx=ln(((π+ϕ)/π)) ϕ::= golen ratio...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\mathrm{nice}\:\:\mathrm{calculus}... \\ $$$${prove}\:\:{that}::\: \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+\varphi^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }\:{dx}={ln}\left(\frac{\pi+\varphi}{\pi}\right) \\ $$$$\varphi::=\:\:{golen}\:{ratio}... \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 09/Jan/21

I(a)=∫_0 ^∞ ((log(1+a^2 x^2 ))/(1+π^2 x^2 ))dx ⇒I′(a)=∫_0 ^∞ ((2ax^2 )/((1+a^2 x^2 )(1+π^2 x^2 )))dx =((2a)/(a^2 −π^2 ))∫_0 ^∞ (1/(1+π^2 x^2 ))−(1/(1+a^2 x^2 ))dx =((2a)/(a^2 −π^2 ))((1/π).(π/2)−(1/a).(π/2))=(1/(π+a)) I(a)=log(π+a)+C C=−log(π) when a=0 I(a)=log(1+(a/π)) ⇒I(ϕ)=log(1+(ϕ/π))

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{log}\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }{dx}\:\Rightarrow{I}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{ax}^{\mathrm{2}} }{\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} −\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} −\pi^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\pi}.\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{a}}.\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\pi+{a}} \\ $$$${I}\left({a}\right)={log}\left(\pi+{a}\right)+{C} \\ $$$${C}=−{log}\left(\pi\right)\:\:\:\:{when}\:{a}=\mathrm{0} \\ $$$${I}\left({a}\right)={log}\left(\mathrm{1}+\frac{{a}}{\pi}\right)\:\Rightarrow{I}\left(\varphi\right)={log}\left(\mathrm{1}+\frac{\varphi}{\pi}\right) \\ $$

Commented by mnjuly1970 last updated on 09/Jan/21

short solution but very nice.thank you...

$$\:{short}\:{solution}\:{but} \\ $$$${very}\:{nice}.{thank}\:{you}... \\ $$

Commented by Dwaipayan Shikari last updated on 09/Jan/21

With pleasure sir!

$${With}\:{pleasure}\:{sir}! \\ $$

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